A square hole is punched out of a circular lamina, the diagonal of the square coinciding with the radius of circle. If ‘a’ be the diameter
of the circle, the distance of the centre of mass of reminder from previous centre of mass. 

Before deleting the square, set the mass of the circular lamina to m.
removing the square lamina's mass

$=\frac{\mathrm{M}}{{\mathrm{\pi R}}^2}\times \left(\frac{1}{2}{\mathrm{R}}^2\right)=\frac{\mathrm{M}}{2\mathrm{\pi }}$

If an equal and symmetrical man is removed from the cone, the cone won't shift in the vertical direction.
let cone of remaining lamina=x

$\frac{\left({\mathrm{M}}_{\mathrm{remainingLamina}}\mathrm{x}\right)+\left({\mathrm{M}}_{\mathrm{squareLamina}}\right){\mathrm{x}}_{\mathrm{squarelamina}}}{\mathrm{total} \mathrm{mass}}=$without eliminating the square, a cone of circular lamina

$\frac{\left(\mathrm{M}-{\displaystyle \frac{\mathrm{M}}{2\mathrm{\pi }}}\right)\mathrm{x}+\left({\displaystyle \frac{\mathrm{M}}{2\mathrm{\pi }}}\right)\left({\displaystyle \frac{-\mathrm{a}}{4}}\right)}{\mathrm{M}}=0$

$\frac{\mathrm{Mx}\left(1-{\displaystyle \frac{1}{2\mathrm{\pi }}}\right)+\left({\displaystyle \frac{\mathrm{M}}{2\mathrm{\pi }}}\right)\left({\displaystyle \frac{-\mathrm{a}}{4}}\right)}{\mathrm{M}}=0$

$\mathrm{x}\left(1-\frac{1}{2\mathrm{\pi }}\right)+\left(\frac{1}{2\mathrm{\pi }}\right)\left(\frac{-\mathrm{a}}{4}\right)=0$

$\mathrm{x}\left(1-\frac{1}{2\mathrm{\pi }}\right)=\frac{\mathrm{a}}{8\mathrm{\pi }}$

$\mathrm{x}=\frac{\mathrm{a}}{8\mathrm{\pi }-4}=\frac{\mathrm{a}}{4(2\mathrm{\pi }-1)}$

Hence the correct answer is $\frac{a}{4(2\mathrm{\pi }-1)}.$